textbook states...

so
v(t) = 2 + 5 sin(2pi60t) - 3 cos(4pi60t + 45deg)
i(t) = 1.5 + 2 cos(2pi60t + 20deg) + 1.1cos(4pi60t)

I assume I cannot just multiply these together then take the integral.
question is do I have to simplify them first?

 

If the equation says to multiply them out and take the integral then you can multiply them out and take the integral

I can't think of a simpler method than just grinding away at that... the simplification can come when you get to the integral step.

Just remember that the average of a cosine or sine over one period, no matter the phase or anything else, is zero.
A sinusoid needs to be multiplied by another sinusoid to contribute to the average.
So, cos(x)^2 or sin(x)^2 or cos(x)sin(x + a) will contribute, sin(x) will not.

 

actually posted a correction to the problem.. here is the new one...

so let me get this straight. if a sinusoid needs to be multiplied by another sinusoid to contribute to the average then none of these sinusoids are eligible to contribute?

so were left with v(t) = 2 and i(t) = 1.5?
and with the period = 60
1/60[(2*1.5)*60] = 3w

this cant be that easy

 

Hm?

You take the integral over a period after you multiply it out.

(a + b)(c + d) = ac + ad + bc + bd

There are plenty of terms that can contribute.

Edit:
Hm, I don't remember whether sinusoids of different frequencies multiplied together have a non-zero average...

 

v(t) = 2 + 5cos(2pi60t) - 3cos(4pi60t + 45deg)

2 = voltage peak to peak
2pi60t = frequency
4pi60t = frequency
45deg = phase shift

are these statements correct?

 

Not entirely...

2pi60t and 4pi60t are both phase... 45 degrees is phase shift.

For cos(wt + x) you have w being angular frequency, wt being phase and x being phase shift.

w is in radians, and w = 2pi*f, where f is in hertz.

So for 2pi60t the frequency is 60 Hz.

2 is the voltage DC component.

 

(a + b)(c + d) = ac + ad + bc + bd

for this statement... i understand distribution properties but I am still confused

which values should I multiply??? just the ones outside the parenthesis? or the phases and phase shifts within the parenthesis?

 

(a + b)(c + d) = ac + ad + bc + bd

v(t) = 2 + 5 cos(2pi60t) - 3 cos(4pi60t + 45deg)
i(t) = 1.5 + 2 cos(2pi60t + 20deg) + 1.1cos(4pi60t - 20deg)

so your saying it should be set up as follows?

[(2) + (5 cos(2pi60t)) - (3 cos(4pi60t + 45deg))]*[(1.5) + (2 cos(2pi60t + 20deg)) + (1.1cos(4pi60t - 20deg))]

 

Yeah that's what would you get you there.

You'll need to use these...
http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

Based on that it does look like the product of two different frequencies has zero average.

 

based on that formula...

2*1.5=3
2*(2 cos(2pi60t + 20deg))=4 cos(2pi60t + 20deg)
2*(1.1cos(4pi60t - 20deg))=2.2cos(4pi60t - 20deg)

(5 cos(2pi60t))*1.5=7.5 cos(2pi60t)
(5 cos(2pi60t))*(2 cos(2pi60t + 20deg))=10....???

(cos(2pi60t))*(cos(2pi60t+20deg)) = [cos(2pi60t-2pi60t+20deg)+cos(2pi60t+2pi60t+20deg)]/2

=[cos(0+20deg)+cos(4pi60t+20deg)]/2
=i dont know how to do this with the + 20deg in there and im not sure how to add the 2pi60t's together

 

allright I am determined to figure this problem out. I have come across another equation that I think will be easier once I plug in the values

Note that the average power is the sum of the powers at the frequencies in the Fourier series.

so now back to the original problem

I understand that with each frequency there is going to be a separate average power value and at the end, I will add them all together.

So for this problem there are three groups of frequencies
v(t) @ 2 and i(t) @ 1.5
V(t) @ 5cos(2pi60t) and i(t) @ 2cos(2pi60t+20deg)
v(t) @ -3cos(4pi60t+45deg) and i(t) @ 1.1cos(4pi60t-20deg)

I also understand the concept and evaluation of Sigma or the summation, however I am really having trouble trying to plug these values into the right spot in this formula